Question: You have found the following ages (in years) of all 6 lizards at your local zoo: $ 2,\enspace 1,\enspace 1,\enspace 1,\enspace 3,\enspace 3$ What is the average age of the lizards at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 lizards at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{2 + 1 + 1 + 1 + 3 + 3}{{6}} = {1.8\text{ years old}} $ Find the squared deviations from the mean for each lizard. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $2$ years $0.2$ years $0.04$ years $^2$ $1$ year $-0.8$ years $0.64$ years $^2$ $1$ year $-0.8$ years $0.64$ years $^2$ $1$ year $-0.8$ years $0.64$ years $^2$ $3$ years $1.2$ years $1.44$ years $^2$ $3$ years $1.2$ years $1.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {0.64} + {0.64} + {0.64} + {1.44} + {1.44}} {{6}} $ $ {\sigma^2} = \dfrac{{4.84}}{{6}} = {0.81\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{0.81\text{ years}^2}} = {0.9\text{ years}} $ The average lizard at the zoo is 1.8 years old. There is a standard deviation of 0.9 years.